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3 years ago

Please help me on the picture above. I will give points for people who help.

Mathematics

2 answers:

Jobisdone [24]3 years ago

6 0

Answer:

I believe the one you selected is correct. A negative plus a negative makes a positive meaning you start at -2 and end with +2

vovikov84 [41]3 years ago

6 0

Answer:

Answer choice b. is correct. You have it right.

Step-by-step explanation:

You go left 2 because of the negative sign. Since you are subtracting a negative number, you add the opposite, so you add 4.

-2-(-4)

-2+4

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3 years ago

A phone manufacturer wants to compete in the touch screen phone market. He understands that the lead product has a battery life

alisha [4.7K]

Answer:

The null hypothesis is $H_0: \mu \leq 10$

The alternative hypothesis is $H_1: \mu > 10$

b-1) The value of the test statistic is t = 1.86.

b-2) The p-value is of 0.0348.

Step-by-step explanation:

Question a:

Test if the battery life is more than twice of 5 hours:

Twice of 5 hours = 5*2 = 10 hours.

At the null hypothesis, we test if the battery life is of 10 hours or less, than is:

$H_0: \mu \leq 10$

At the alternative hypothesis, we test if the battery life is of more than 10 hours, that is:

$H_1: \mu > 10$

b-1. Calculate the value of the test statistic.

The test statistic is:

We have the standard deviation for the sample, so the t-distribution is used to solve this question

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

In which X is the sample mean, $\mu$ is the value tested at the null hypothesis, s is the standard deviation and n is the size of the sample.

10 is tested at the null hypothesis:

This means that $\mu = 10$

In order to test the claim, a researcher samples 45 units of the new phone and finds that the sample battery life averages 10.5 hours with a sample standard deviation of 1.8 hours.

This means that $n = 45, X = 10.5, s = 1.8$

Then

$t = \frac{X - \mu}{\frac{s}{\sqrt{n}}}$

$t = \frac{10.5 - 10}{\frac{1.8}{\sqrt{45}}}$

$t = 1.86$

The value of the test statistic is t = 1.86.

b-2. Find the p-value.

Testing if the mean is more than a value, so a right-tailed test.

Sample of 45, so 45 - 1 = 44 degrees of freedom.

Test statistic t = 1.86.

Using a t-distribution calculator, the p-value is of 0.0348.

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3 years ago

Scores on an exam follow an approximately Normal distribution with a mean of 76.4 and a standard deviation of 6.1 points. What p

klasskru [66]

Answer:

99.89% of students scored below 95 points.

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 76.4, \sigma = 6.1$

What percent of students scored below 95 points?

This is the pvalue of Z when X = 95. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{95 - 76.4}{6.1}$

$Z = 3.05$

$Z = 3.05$ has a pvalue of 0.9989.

99.89% of students scored below 95 points.

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3 years ago